Random Networks. Complex Networks, CSYS/MATH 303, Spring, Prof. Peter Dodds

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1 Complex Networks, CSYS/MATH 303, Spring, 2010 Prof. Peter Dodds Department of Mathematics & Statistics Center for Complex Systems Vermont Advanced Computing Center University of Vermont Licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License. Frame 1/89

2 Outline Frame 2/89

3 Random networks Pure, abstract random networks: Consider set of all networks with N labelled nodes and m edges. Standard random network = randomly chosen network from this set. To be clear: each network is equally probable. Sometimes equiprobability is a good assumption, but it is always an assumption. Known as Erdős-Rényi random networks or ER graphs. Frame 4/89

4 Random networks Some features: Number of possible edges: ( ) N N(N 1) 0 m = 2 2 Given m edges, there are ( ( N ) 2) m different possible networks. Crazy factorial explosion for 1 m ( N 2). Limit of m = 0: empty graph. Limit of m = ( N 2) : complete or fully-connected graph. Real world: links are usually costly so real networks are almost always sparse. Frame 5/89

5 Random networks standard random networks: Given N and m. Two probablistic methods (we ll see a third later on) 1. Connect each of the ( N 2) pairs with appropriate probability p. Useful for theoretical work. 2. Take N nodes and add exactly m links by selecting edges without replacement. Algorithm: Randomly choose a pair of nodes i and j, i j, and connect if unconnected; repeat until all m edges are allocated. Best for adding relatively small numbers of links (most cases). 1 and 2 are effectively equivalent for large N. Frame 7/89

6 Random networks A few more things: For method 1, # links is probablistic: ( ) N m = p = p 1 N(N 1) 2 2 So the expected or average degree is k = 2 m N = 2 N p 1 2 N(N 1) = 2 N p 1 N(N 1) = p(n 1). 2 Which is what it should be... If we keep k constant then p 1/N 0 as N. Frame 8/89

7 Random networks: examples Next slides: Example realizations of random networks N = 500 Vary m, the number of edges from 100 to Average degree k runs from 0.4 to 4. Look at full network plus the largest component. Frame 10/89

8 Random networks: examples entire network: largest component: N = 500, number of edges m = 100 average degree k = 0.4 Frame 11/89

9 Random networks: examples entire network: largest component: N = 500, number of edges m = 200 average degree k = 0.8 Frame 12/89

10 Random networks: examples entire network: largest component: N = 500, number of edges m = 230 average degree k = 0.92 Frame 13/89

11 Random networks: examples entire network: largest component: N = 500, number of edges m = 240 average degree k = 0.96 Frame 14/89

12 Random networks: examples entire network: largest component: N = 500, number of edges m = 250 average degree k = 1 Frame 15/89

13 Random networks: examples entire network: largest component: N = 500, number of edges m = 260 average degree k = 1.04 Frame 16/89

14 Random networks: examples entire network: largest component: N = 500, number of edges m = 280 average degree k = 1.12 Frame 17/89

15 Random networks: examples entire network: largest component: N = 500, number of edges m = 300 average degree k = 1.2 Frame 18/89

16 Random networks: examples entire network: largest component: N = 500, number of edges m = 500 average degree k = 2 Frame 19/89

17 Random networks: examples entire network: largest component: N = 500, number of edges m = 1000 average degree k = 4 Frame 20/89

18 Random networks: examples for N=500 m = 100 k = 0.4 m = 200 k = 0.8 m = 230 k = 0.92 m = 240 k = 0.96 m = 250 k = 1 m = 260 k = 1.04 m = 280 k = 1.12 m = 300 k = 1.2 m = 500 k = 2 m = 1000 k = 4 Frame 21/89

19 Random networks: largest components m = 100 k = 0.4 m = 200 k = 0.8 m = 230 k = 0.92 m = 240 k = 0.96 m = 250 k = 1 m = 260 k = 1.04 m = 280 k = 1.12 m = 300 k = 1.2 m = 500 k = 2 m = 1000 k = 4 Frame 22/89

20 Random networks: examples for N=500 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 Frame 23/89

21 Random networks: largest components m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 m = 250 k = 1 Frame 24/89

22 Random networks : For method 1, what is the clustering coefficient for a finite network? Consider triangle/triple clustering coefficient (Newman [1] ): C 2 = 3 #triangles #triples Recall: C 2 = probability that two nodes are connected given they have a friend in common. For standard random networks, we have simply that C 2 = p. Frame 26/89

23 Random networks : So for large random networks (N ), clustering drops to zero. Key structural feature of random networks is that they locally look like branching networks (no loops). Frame 27/89

24 Random networks Degree distribution: Recall p k = probability that a randomly selected node has degree k. Consider method 1 for constructing random networks: each possible link is realized with probability p. Now consider one node: there are N 1 choose k ways the node can be connected to k of the other N 1 nodes. Each connection occurs with probability p, each non-connection with probability (1 p). Therefore have a binomial distribution: ( ) N 1 P(k; p, N) = p k (1 p) N 1 k. k Frame 29/89

25 Random networks Limiting form of P(k; p, N): Our degree distribution: P(k; p, N) = ( ) N 1 k p k (1 p) N 1 k. What happens as N? We must end up with the normal distribution right? If p is fixed, then we would end up with a Gaussian with average degree k pn. But we want to keep k fixed... So examine limit of P(k; p, N) when p 0 and N with k = p(n 1) = constant. Frame 30/89

26 Limiting form of P(k; p, N): Substitute p = k N 1 into P(k; p, N) and hold k fixed: = ( N 1 P(k; p, N) = k ) ( k N 1 (N 1)! k k = k!(n 1 k)! (N 1) k (N 1)(N 2) (N k) k! N k (1 1 N ) (1 N k ) k k k! N k (1 1 N )k ) k ( 1 k ) N 1 k N 1 ( 1 k ) N 1 k N 1 k k (N 1) k ( 1 k ) N 1 k N 1 ( 1 k ) N 1 k N 1 Frame 31/89

27 Limiting form of P(k; p, N): We are now here: P(k; p, N) k k k! ( 1 k ) N 1 k N 1 Now use the excellent result: ( lim 1 + x ) n = e x. n n (Use l Hôpital s rule to prove.) Identifying n = N 1 and x = k : ( P(k; k ) k k k! e k 1 k ) k k k N 1 k! e k This is a Poisson distribution ( ) with mean k. Frame 32/89

28 General random networks So... standard random networks have a Poisson degree distribution Generalize to arbitrary degree distribution P k. Also known as the configuration model [1]. Can generalize construction method from ER random networks. Assign each node a weight w from some distribution P w and form links with probability P(link between i and j) w i w j. But we ll be more interested in 1. Randomly wiring up (and rewiring) already existing nodes with fixed degrees. 2. Examining mechanisms that lead to networks with certain degree distributions. Frame 34/89

29 Random networks: examples Coming up: Example realizations of random networks with power law degree distributions: N = P k k γ for k 1. Set P 0 = 0 (no isolated nodes). Vary exponent γ between 2.10 and Again, look at full network plus the largest component. Apart from degree distribution, wiring is random. Frame 35/89

30 Random networks: examples for N=1000 γ = 2.1 k = γ = 2.19 k = γ = 2.28 k = γ = 2.37 k = γ = 2.46 k = γ = 2.55 k = γ = 2.64 k = 1.6 γ = 2.73 k = γ = 2.82 k = γ = 2.91 k = 1.49 Frame 36/89

31 Random networks: largest components γ = 2.1 k = γ = 2.19 k = γ = 2.28 k = γ = 2.37 k = γ = 2.46 k = γ = 2.55 k = γ = 2.64 k = 1.6 γ = 2.73 k = γ = 2.82 k = γ = 2.91 k = 1.49 Frame 37/89

32 Poisson basics: Normalization: we must have Checking: P(k; k ) = 1 k=0 P(k; k ) = k=0 k=0 = e k k=0 k k k! e k k k k! = e k e k = 1 Frame 38/89

33 Poisson basics: Mean degree: we must have k = kp(k; k ). Checking: k=0 kp(k; k ) = k k k k! e k k=0 = k e k = e k k=1 k=0 = k e k i=0 k i i! k=1 k k (k 1)! k k 1 (k 1)! = k e k e k = k We ll get to a better way of doing this... Frame 39/89

34 Poisson basics: The variance of degree distributions for random networks turns out to be very important. Use calculation similar to one for finding k to find the second moment: Variance is then k 2 = k 2 + k. σ 2 = k 2 k 2 = k 2 + k k 2 = k. So standard deviation σ is equal to k. Note: This is a special property of Poisson distribution and can trip us up... Frame 40/89

35 The edge-degree distribution: The degree distribution P k is fundamental for our description of many complex networks Again: P k is the degree of randomly chosen node. A second very important distribution arises from choosing randomly on edges rather than on nodes. Define Q k to be the probability the node at a random end of a randomly chosen edge has degree k. Now choosing nodes based on their degree (i.e., size): Q k kp k Normalized form: Q k = kp k k =0 k = kp k P k k. Frame 41/89

36 The edge-degree distribution: For random networks, Q k is also the probability that a friend (neighbor) of a random node has k friends. Useful variant on Q k : R k = probability that a friend of a random node has k other friends. (k + 1)P k+1 R k = k =0 (k + 1)P k +1 = (k + 1)P k+1 k Equivalent to friend having degree k + 1. Natural question: what s the expected number of other friends that one friend has? Frame 42/89

37 The edge-degree distribution: Given R k is the probability that a friend has k other friends, then the average number of friends other friends is k R = kr k = = 1 k k=0 = 1 k k=0 k (k + 1)P k+1 k k(k + 1)P k+1 k=1 ( (k + 1) 2 (k + 1) ) P k+1 k=1 (where we have sneakily matched up indices) = 1 k (j 2 j)p j (using j = k+1) j=0 = 1 ( k 2 k ) k Frame 43/89

38 The edge-degree distribution: Note: our result, k R = 1 ( k k 2 k ), is true for all random networks, independent of degree distribution. For standard random networks, recall k 2 = k 2 + k. Therefore: k R = 1 ( ) k 2 + k k = k k Again, neatness of results is a special property of the Poisson distribution. So friends on average have k other friends, and k + 1 total friends... Frame 44/89

39 Two reasons why this matters Reason #1: Average # friends of friends per node is k 2 = k k R = k 1 ( k 2 k ) = k 2 k. k Key: Average depends on the 1st and 2nd moments of P k and not just the 1st moment. Three peculiarities: 1. We might guess k 2 = k ( k 1) but it s actually k(k 1). 2. If P k has a large second moment, then k 2 will be big. (e.g., in the case of a power-law distribution) 3. Your friends are different to you... Frame 45/89

40 Two reasons why this matters More on peculiarity #3: A node s average # of friends: k Friend s average # of friends: k 2 k Comparison: k 2 k = k k 2 k 2 = k σ2 + k 2 k 2 ) = k (1 + σ2 k 2 k So only if everyone has the same degree (variance= σ 2 = 0) can a node be the same as its friends. Intuition: for random networks, the more connected a node, the more likely it is to be chosen as a friend. Frame 46/89

41 Two reasons why this matters (Big) Reason #2: k R is key to understanding how well random networks are connected together. e.g., we d like to know what s the size of the largest component within a network. As N, does our network have a giant component? Defn: = connected subnetwork of nodes such that path between each pair of nodes in the subnetwork, and no node outside of the subnetwork is connected to it. Defn: Giant component = component that comprises a non-zero fraction of a network as N. Note: = Cluster Frame 47/89

42 of random networks Giant component: A giant component exists if when we follow a random edge, we are likely to hit a node with at least 1 other outgoing edge. Equivalently, expect exponential growth in node number as we move out from a random node. All of this is the same as requiring k R > 1. Giant component condition (or percolation condition): k R = k 2 k k > 1 Again, see that the second moment is an essential part of the story. Equivalent statement: k 2 > 2 k Frame 49/89

43 Giant component Standard random networks: Recall k 2 = k 2 + k. Condition for giant component: k R = k 2 k k = k 2 + k k k = k Therefore when k > 1, standard random networks have a giant component. When k < 1, all components are finite. Fine example of a continuous phase transition ( ). We say k = 1 marks the critical point of the system. Frame 50/89

44 Giant component Random networks with skewed P k : e.g, if P k = ck γ with 2 < γ < 3 then k 2 = c x 3 γ x=0 x=0 k 2 k γ k=0 x 2 γ dx = (> k ). So giant component always exists for these kinds of networks. Cutoff scaling is k 3 : if γ > 3 then we have to look harder at k R. Frame 51/89

45 Giant component And how big is the largest component? Define S 1 as the size of the largest component. Consider an infinite ER random network with average degree k. Let s find S 1 with a back-of-the-envelope argument. Define δ as the probability that a randomly chosen node does not belong to the largest component. Simple connection: δ = 1 S 1. Dirty trick: If a randomly chosen node is not part of the largest component, then none of its neighbors are. So δ = P k δ k k=0 Substitute in Poisson distribution... Frame 52/89

46 Giant component Carrying on: δ = P k δ k = k=0 k=0 = e k ( k δ) k k! k=0 k k k! e k δ k = e k e k δ = e k (1 δ). Now substitute in δ = 1 S 1 and rearrange to obtain: S 1 = 1 e k S 1. Frame 53/89

47 Giant component We can figure out some limits and details for S 1 = 1 e k S 1. First, we can write k in terms of S 1 : As k 0, S 1 0. As k, S 1 1. k = 1 S 1 ln 1 1 S 1. Notice that at k = 1, the critical point, S 1 = 0. Only solvable for S > 0 when k > 1. Really a transcritical bifurcation [2]. Frame 54/89

48 Giant component S k Frame 55/89

49 Giant component Turns out we were lucky... Our dirty trick only works for ER random networks. The problem: We assumed that neighbors have the same probability δ of belonging to the largest component. But we know our friends are different from us... Works for ER random networks because k = k R. We need a separate probability δ for the chance that a node at the end of a random edge is part of the largest component. We can do this but we need to enhance our toolkit with functionology... [3] (Well, not really but it s fun and we get all sorts of other things...) Frame 56/89

50 functions Idea: Given a sequence a 0, a 1, a 2,..., associate each element with a distinct function or other mathematical object. Well-chosen functions allow us to manipulate sequences and retrieve sequence elements. Definition: The generating function (g.f.) for a sequence {a n } is F (x) = a n x n. n=0 Roughly: transforms a vector in R into a function defined on R 1. Related to Fourier, Laplace, Mellin,... Frame 58/89

51 Simple example Rolling dice: p ( ) k = Pr(throwing a k) = 1/6 where k = 1, 2,..., 6. F ( ) (x) = 6 p k x k = 1 6 (x + x 2 + x 3 + x 4 + x 5 + x 6 ). k=1 We ll come back to this simple example as we derive various delicious properties of generating functions. Frame 59/89

52 Example Take a degree distribution with exponential decay: where c = 1 e λ. P k = ce λk The generating function for this distribution is F(x) = P k x k = k=0 ce λk x k = k=0 Notice that F (1) = c/(1 e λ ) = 1. c 1 xe λ. For probability distributions, we must always have F(1) = 1 since F (1) = P k 1 k = k=0 P k = 1. k=0 Frame 60/89

53 Properties of generating functions Average degree: k = kp k = kp k x k 1 k=0 k=0 x=1 = d dx F (x) = F (1) x=1 In general, many calculations become simple, if a little abstract. For our exponential example: F (x) = (1 e λ )e λ (1 xe λ ) 2. So: k = F (1) = e λ (1 e λ ). Frame 62/89

54 Properties of generating functions Useful pieces for probability distributions: Normalization: First moment: Higher moments: k n = F (1) = 1 k = F (1) ( x d dx ) n F (x) k th element of sequence (general): P k = 1 d k k! dx k F (x) x=1 x=0 Frame 63/89

55 Edge-degree distribution Recall our condition for a giant component: k R = k 2 k k > 1. Let s rëexpress our condition in terms of generating functions. We first need the g.f. for R k. We ll now use this notation: F P (x) is the g.f. for P k. F R (x) is the g.f. for R k. Condition in terms of g.f. is: k R = F R (1) > 1. Now find how F R is related to F P... Frame 65/89

56 Edge-degree distribution We have F R (x) = R k x k = k=0 k=0 (k + 1)P k+1 x k. k Shift index to j = k + 1 and pull out 1 k : = 1 d k dx F R (x) = 1 k j=1 jp j x j 1 = 1 k j=1 P j d dx x j P j x j = 1 d k dx (F P(x) P 0 ) = 1 k F P (x). j=1 Finally, since k = F P (1), F R (x) = F P (x) F P (1) Frame 66/89

57 Edge-degree distribution Recall giant component condition is k R = F R (1) > 1. Since we have F R (x) = F P (x)/f P (1), F R (x) = F P (x) F P (1). Setting x = 1, our condition becomes F P (1) F P (1) > 1 Frame 67/89

58 Size distributions To figure out the size of the largest component (S 1 ), we need more resolution on component sizes. : π n = probability that a random node belongs to a finite component of size n <. ρ n = probability a random link leads to a finite subcomponent of size n <. Local-global connection: P k, R k π n, ρ n neighbors components Frame 69/89

59 Size distributions G.f. s for component size distributions: F π (x) = π n x n and F ρ (x) = ρ n x n n=0 n=0 The largest component: Subtle key: F π (1) is the probability that a node belongs to a finite component. Therefore: S 1 = 1 F π (1). Our mission, which we accept: Find the four generating functions F P, F R, F π, and F ρ. Frame 70/89

60 we ll need for g.f. s Sneaky Result 1: Consider two random variables U and V whose values may be 0, 1, 2,... Write probability distributions as U k and V k and g.f. s as F U and F V. SR1: If a third random variable is defined as then U W = V (i) with each V (i) d = V i=1 F W (x) = F U (F V (x)) Frame 72/89

61 Proof of SR1: Write probability that variable W has value k as W k. W k = U j Pr(sum of j draws of variable V = k) j=0 = j=0 U j {i 1,i 2,...,i j } i 1 +i i j =k V i1 V i2 V ij F W (x) = W k x k = k=0 k=0 j=0 U j {i 1,i 2,...,i j } i 1 +i i j =k V i1 V i2 V ij x k = U j j=0 k=0 {i 1,i 2,...,i j } i 1 +i i j =k V i1 x i 1 V i2 x i2 V ij x i j Frame 73/89

62 Proof of SR1: With some concentration, observe: F W (x) = U j j=0 k=0 {i 1,i 2,...,i j } i 1 +i i j =k V i1 x i 1 V i2 x i2 V ij x i j }{{} ( x k piece of i =0 V i xi ) j }{{} ( i =0 V i xi ) j = (F V (x)) j = U j (F V (x)) j j=0 = F U (F V (x)) Frame 74/89

63 we ll need for g.f. s Sneaky Result 2: Start with a random variable U with distribution U k (k = 0, 1, 2,... ) SR2: If a second random variable is defined as V = U + 1 then F V (x) = xf U (x) Reason: V k = U k 1 for k 1 and V 0 = 0. F V (x) = = x V k x k = k=0 U k 1 x k k=1 U j x j = xf U (x). j=0 Frame 75/89

64 we ll need for g.f. s Generalization of SR2: (1) If V = U + i then (2) If V = U i then F V (x) = x i F U (x). F V (x) = x i F U (x) = x i U k x k k=0 Frame 76/89

65 Connecting generating functions Goal: figure out forms of the component generating functions, F π and F ρ. π n = probability that a random node belongs to a finite component of size n = k=0 ( sum of sizes of subcomponents P k Pr at end of k random links = n 1 Therefore: F π (x) = }{{} x F P (F ρ (x)) }{{} SR2 SR1 ) Extra factor of x accounts for random node itself. Frame 78/89

66 Connecting generating functions ρ n = probability that a random link leads to a finite subcomponent of size n. Invoke one step of recursion: ρ n = probability that in following a random edge, the outgoing edges of the node reached lead to finite subcomponents of combined size n 1, = k=0 ( sum of sizes of subcomponents R k Pr at end of k random links = n 1 Therefore: F ρ (x) = }{{} x F R (F ρ (x)) }{{} Again, extra factor of x accounts for random node itself. SR2 SR1 ) Frame 79/89

67 Connecting generating functions We now have two functional equations connecting our generating functions: F π (x) = xf P (F ρ (x)) and Taking stock: We know F P (x) and F R (x) = F P (x)/f P (1). F ρ (x) = xf R (F ρ (x)) We first untangle the second equation to find F ρ We can do this because it only involves F ρ and F R. The first equation then immediately gives us F π in terms of F ρ and F R. Frame 80/89

68 Remembering vaguely what we are doing: Finding F π to obtain the fractional size of the largest component S 1 = 1 F π (1). Set x = 1 in our two equations: F π (1) = F P (F ρ (1)) and F ρ (1) = F R (F ρ (1)) Solve second equation numerically for F ρ (1). Plug F ρ (1) into first equation to obtain F π (1). Frame 81/89

69 Example: Standard random graphs. We can show F P (x) = e k (1 x) F R (x) = F P (x)/f P (1) = e k (1 x) /e k (1 x ) x =1 = e k (1 x) = F P (x)...aha! RHS s of our two equations are the same. So F π (x) = F ρ (x) = xf R (F ρ (x)) = xf R (F π (x)) Why our dirty (but wrong) trick worked earlier... Frame 82/89

70 We are down to F π (x) = xf R (F π (x)) and F R (x) = e k (1 x). F π (x) = xe k (1 Fπ(x)) We re first after S 1 = 1 F π (1) so set x = 1 and replace F π (1) by 1 S 1 : 1 S 1 = e k S 1 Or: k = 1 1 ln S 1 1 S 1 S k Just as we found with our dirty trick... Again, we (usually) have to resort to numerics Frame 83/89

71 Average component size Next: find average size of finite components n. Using standard G.F. result: n = F π(1). Try to avoid finding F π (x)... Starting from F π (x) = xf P (F ρ (x)), we differentiate: F π(x) = F P (F ρ (x)) + xf ρ(x)f P (F ρ(x)) While F ρ (x) = xf R (F ρ (x)) gives F ρ(x) = F R (F ρ (x)) + xf ρ(x)f R (F ρ(x)) Now set x = 1 in both equations. We solve the second equation for F ρ(1) (we must already have F ρ (1)). Plug F ρ(1) and F ρ (1) into first equation to find F π(1). Frame 85/89

72 Average component size Example: Standard random graphs. Use fact that F P = F R and F π = F ρ. Two differentiated equations reduce to only one: F π(x) = F P (F π (x)) + xf π(x)f P (F π(x)) Rearrange: F π(x) = F P (F π (x)) 1 xf P (F π(x)) Simplify denominator using F P (x) = k F P(x) Replace F P (F π (x)) using F π (x) = xf P (F π (x)). Set x = 1 and replace F π (1) with 1 S 1. End result: n = F π(1) = (1 S 1 ) 1 k (1 S 1 ) Frame 86/89

73 Average component size Our result for standard random networks: n = F π(1) = (1 S 1 ) 1 k (1 S 1 ) Recall that k = 1 is the critical value of average degree for standard random networks. Look at what happens when we increase k to 1 from below. We have S 1 = 0 for all k < 1 so n = 1 1 k This blows up as k 1. Reason: we have a power law distribution of component sizes at k = 1. Typical critical point behavior... Frame 87/89

74 Average component size Limits of k = 0 and make sense for n = F π(1) = As k 0, S 1 = 0, and n 1. All nodes are isolated. (1 S 1 ) 1 k (1 S 1 ) As k, S 1 1 and n 0. No nodes are outside of the giant component. Extra on largest component size: For k = 1, S 1 N 2/3. For k < 1, S 1 log N. Frame 88/89

75 I [1] M. E. J. Newman. The structure and function of complex networks. SIAM Review, 45(2): , pdf ( ) [2] S. H. Strogatz. Nonlinear Dynamics and Chaos. Addison Wesley, Reading, Massachusetts, [3] H. S. Wilf. functionology. A K Peters, Natick, MA, 3rd edition, pdf ( ) Frame 89/89